Notice that I did NOT just get "3x^2- 3" and "3y^2- 12"! I said that, in order to have a critical point, we must have and from that I got the equations "3x^2- 3= 0" and "3y^2- 12= 0". Yes, and satisfy those equations. Since any pairing makes the gradient 0, the critical points are (1, 2), (1, -2), (-1, 2), and (-1, -2).